Calculate the pooled estimate of variance

[chwala]

Homework Statement

See attached

Relevant Equations

stats

OK, Let me attempt part (i), first,
Here we have;
$s^2_p$=$\dfrac{ (n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$

$s^2_p$=$\dfrac{ (7-1)0.63953+(7-1)0.6148}{7+7-2}$

$s^2_p$=$\dfrac{3.83718+3.6888}{12}$

$s^2_p$=$\dfrac{3.83718+3.6888}{12}$

$s^2_p$=$\dfrac{7.52598}{12}$

$s^2_p$=$0.627165$ its located between the two original variances... correct? i do not have markscheme nor solutions...

* Reading on this topic now...the literature is really confusing on the so called population data (presumably all the n values in a given data set) and sample data...

 

[nuuskur]

i) Your sample sizes are uniform, so there's no need for weighted averages here. Find the variance estimates for both populations and then take their arithmetic mean.

 

[chwala]

i) Your sample sizes are uniform, so there's no need for weighted averages here. Find the variance estimates for both populations and then take their arithmetic mean.

Ok mate i.e $\dfrac {0.63953+0.6148}{2}$= $\dfrac {1.25433}{2}=0.627165$ .

For part (ii), Let, $μ_1$ and$μ_2$ be the mean for boys and girls respectively, then
We want to test the hypothesis; as per the question...

$H_0$: $μ_1$=$μ_2$ - Mean weight of boys is equal to mean weight of girls.
$H_A$: $μ_1$<$μ_2$ - Mean weight of boys is less than the mean weight of girls.

Using the t-statistic and also considering that dof =$12$and $α=0.05$ then it follows that the critical value = $-1.782$
We shall therefore have,
$t$=$\dfrac {2.5429-3.18571}{\sqrt{(0.627165^2(\frac {1}{6}+\frac {1}{6})}}$=$\dfrac {2.5429-3.18571}{0.256040263}$=$-2.5105 < -1.782$ We thefore Reject the Null hypothesis and accept the Alternative hypothesis.

 

[nuuskur]

Your notation is overloaded. You likely mean $\mu _1$ for boys and $\mu _2$ for girls. Hypotheses are logical negations of one another. This is not the case right now.

 

[chwala]

Your notation is overloaded. You likely mean $\mu _1$ for boys and $\mu _2$ for girls. Hypotheses are logical negations of one another. This is not the case right now.

Amended ...sorry was a bit busy...

 

[chwala]

Your notation is overloaded. You likely mean $\mu _1$ for boys and $\mu _2$ for girls. Hypotheses are logical negations of one another. This is not the case right now.

Is it now correct?